Bourbaki

Exercise 10, § 8

Tag 02HO, permanent. Cite it as [Bourbaki, Tag 02HO].

Spectral Theories, I, p. 195

Bourbaki marks this one as harder than the rest.

Let X be a compact topological space and A a unital Banach subalgebra of C(X)\mathscr{C}(X) for the induced norm.

a) A subset K of X is said to be antisymmetric (relative to A) if every function f∈Af\in A that is real on K is constant on K. Every antisymmetric subset is contained in a maximal antisymmetric subset. A maximal antisymmetric subset is closed. The set R\mathfrak{R} of maximal antisymmetric subsets is a partition of X.

b) Let A⊥A^{\bot} be the orthogonal of A in the Banach space of complex measures on X. Let μ\mu be an extreme point of the unit ball of A⊥A^{\bot}. Now prove that the support of μ\mu is antisymmetric.

c) Let f∈C(X)f\in \mathscr{C}(X). If f∣K∈A∣Kf|K\in A|K for every K∈RK\in \mathfrak{R}, then f∈Af\in A. (Apply b) and the Krein–Milman theorem.) Hence recover the Weierstrass–Stone theorem (TG, X, p. 36, th. 3).

d) A nonempty subset E of X is called a peak (relative to A) if there exists f∈Af\in A such that ∥f∥=1\|f\|= 1 and E is the set of x∈Xx\in X for which f(x)=1f(x) = 1. One may then suppose that ∣f(x)∣<1|f(x)|<1 for x∉Ex\notin E (replace ff by 12(1+f)\frac{1}{2}(1 +f)). Every nonempty countable intersection of peaks is a peak. If E1,E2E_1,E_2 are peaks, then E1∪E2E_1\cup E_2 is a peak. (Let f1,f2∈Af_1, f_2\in A with fi=1f_i= 1 on Ei,∣fi∣<1E_i,|f_i|<1 on X - EiE_i; let gi=14(1−fi)1/3∈Ag_i=\frac{1}{4}(1-f_i)^{1/3}\in A; then 1−g1g2=11-g_1g_2= 1 on E1∪E2,∣1−g1g2∣<1E_1\cup E_2,|1-g_1g_2|<1 on X - (E1∪E2)(E_1\cup E_2).)

e) Let E be an intersection of peaks. Let I be the set of f∈Af\in A vanishing on E. Then f↦f∣Ef\mapsto f|E defines an isometry of A/IA/I onto the algebra A∣EA|E of restrictions to E of functions in A.

f) Let E be a peak. Let E′⊂EE'\subset E be a peak relative to A∣EA|E. Then E′E' is a peak relative to A.

g) Every antisymmetric subset of X is an intersection of peaks. (Use d) and f).) h) For every antisymmetric subset K of X, the algebra A∣KA|K is a closed subalgebra of C(K)\mathscr{C}(K). (Use e) and g).)

i) Let R be the set of real parts of the elements of A. Suppose that X∈RX\in \mathfrak{R} and that R is stable under multiplication. Then X is reduced to a point. (Let x0∈Xx_0\in X. For u∈Ru\in R, there exists a unique f∈Af\in A such that u=R(f)u=\mathscr{R}(f) and f(x0)∈Rf(x_0)\in \mathbf{R}; set N(u)=∥f∥N(u) =\|f\|. Then R is a real Banach space for N; the closed graph theorem proves that multiplication in R is separately continuous, hence continuous. Deduce on S=R+iRS = R +iR a Banach algebra structure. Let p∈Rp\in R with p(x)>0p(x)>0 for every x∈Xx\in X. Now prove that, for every character χ\chi of S, one has R(χ(p))>0\mathscr{R}(\chi (p))>0, and deduce that log(pp)∈R\in R. Assuming X≠{x0}X\not=\{x_0\}, one can construct an f∈Af\in A such that 0<R(f)⩽10<\mathscr{R}(f)\leqslant 1 on X,f(x0)∈R,f(x_0)\in \mathbf{R} and such that ∥f∥\|f\| is arbitrarily large. From the above, there exists V∈AV\in A with ∣V∣2=R(f)|V|^2=\mathscr{R}(f). One has ∥V∥⩽1\|V\|\leqslant 1 therefore N(R(V))⩽2N(\mathscr{R}(V))\leqslant 2, but

1

N(R(V)2)⩾(∥V2+f∥−∣I(V)(x0)2∣) N(\mathscr{R}(V)^2)\geqslant (\|V^2+f\| - |\mathscr{I}(V)(x_0)^2|)

2

is arbitrarily large.)

j) If R is stable under multiplication, then A=C(X)A =\mathscr{C}(X). (Use a), c), h), i).) Consequently, if A≠C(X)A\not=\mathscr{C}(X), there exists u∈Ru\in R such that u2∉Ru^2\notin R.

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Machine translation of the English, by gpt-5.4. Not checked by a person.